Problem: The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives $30$ years; the standard deviation is $4.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a zebra living between $38.8$ and $43.2$ years.
Solution: $30$ $25.6$ $34.4$ $21.2$ $38.8$ $16.8$ $43.2$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $30$ years. We know the standard deviation is $4.4$ years, so one standard deviation below the mean is $25.6$ years and one standard deviation above the mean is $34.4$ years. Two standard deviations below the mean is $21.2$ years and two standard deviations above the mean is $38.8$ years. Three standard deviations below the mean is $16.8$ years and three standard deviations above the mean is $43.2$ years. We are interested in the probability of a zebra living between $38.8$ and $43.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the zebras will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the zebras will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of zebras between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular zebra living between $38.8$ and $43.2$ years is $\color{orange}{2.35\%}$.